The general problem of converting classical expressions to quantum operator ones is in general unsolvable because classical mechanics is an approximation to quantum mechanics and not the other way around. There is always an ambiguity in how to order noncommuting operators. You have to handle it on a case by case basis, and there are a number of "quantization" schemes out there. In general these can lead to different quantum theories which have to be distinguished experimentally.
Anyway, in your case it is probably fine just to use the eigenvalue decomposition:
$$ \frac{1}{x} \to \frac{1}{\hat{x}} \equiv \int \mathrm{d}x\ |x\rangle \frac{1}{x} \langle x |, $$
$$ \frac{1}{p} \to \frac{1}{\hat{p}} \equiv \int \mathrm{d}p\ |p\rangle \frac{1}{p} \langle p |, $$
etc., where $|x\rangle,|p\rangle$ are the orthonormal eigenvectors of position and momentum resp. You can use $\langle x|x'\rangle=\delta(x-x')$ to show that $\frac{1}{\hat{x}}$ has the desired action on position eigenstates. You can also clearly generalise this sort of thing, e.g. $\sqrt{p}\to\sqrt{\hat{p}}\equiv\int \mathrm{d}p\ |p\rangle \sqrt{p} \langle p |$. To give a real example, the following operator, called the resolvent, is very important in quantum scattering theory:
$$ \hat{R}(z) = \frac{1}{\hat{H}-z} = \sum_n \frac{|n\rangle\langle n|}{E_n - z}, $$
where $z$ is a complex number.
You'll have ambiguities if the classical expression is something like $p/x$ or $\sqrt{px}$ or whatever, since $\hat{p}$ and $\hat{x}$ don't commute.
